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My attempt

$\begin{align}S\arctan R & = \frac{1}{i}\log\left(\frac{1}{\sqrt{1+R^2}}+\frac{R}{\sqrt{1+R^2}}i\right)^S\\ &= \frac{1}{i}\log\left(\frac{1}{1+R^2}\right)^\frac{S}{2}\sum_{n\ge 0}{S\choose n}(Ri)^n \end{align}$

But I have no idea how to simplify this expression.

hskimse
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    The left hand side is simple enough isn't it? – NoChance Sep 15 '19 at 13:39
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    @NoChance I just want to know there exists a simplified expression of $\tan(S\arctan R)$ like $\tan(2\arctan A)=\frac{2A}{1-A^2}$, don't be so that mad. – hskimse Sep 15 '19 at 13:44
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    Sorry, I never meant to sound mad at all :) – NoChance Sep 15 '19 at 13:59
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    If you don't want to use a series, like Taylor's, you may express $arctan(x)=arcsin(x/sqrt(1+x^2))$ Ref: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions – NoChance Sep 15 '19 at 14:16

1 Answers1

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Note that the question is essentially equivalent to asking for a simplification of $\tan(\alpha x)$ for a real number $\alpha$, since if we had an expression for $\tan(\alpha x)$ in terms of $\sin(x),\cos(x)$ and $\tan(x)$, then we can simply use the identities $$\sin(\arctan(x))=\frac{x}{\sqrt{x^2+1}},\quad\cos(\arctan(x))=\frac{1}{\sqrt{x^2+1}},\quad\tan(\arctan(x))=x$$ to convert that into a formula for $\tan(\alpha\arctan(x))$. This question becomes one that has been asked many times before, and essentially the answer is that no, there is no nice simplification except for the cases when $\alpha$ is a positive integer. For more details, see this question. Note that the linked question addresses $\cos(\alpha x)$, but the idea is essentially the same for $\tan(\alpha x)$ as well.

YiFan Tey
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